Question: What is the area of the region between the graphs of $f(x)=x^2+2x$ and $g(x)=2x+1$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \dfrac{4}{3}$ (Choice B) B $2$ (Choice C) C $\dfrac{10}{3}$ (Choice D) D $\dfrac{2}{3}$
Solution: Visualizing the area We sketch the graphs of $f$ and $g$ first. ${1}$ ${\llap{-}1}$ ${1}$ ${2}$ ${3}$ $f$ $g$ $y$ $x$ From the graph, it appears that $g(x)\ge f(x)$ between the points where the graphs intersect. From this we are looking to evaluate: $ \int_{a}^{b}\left( g(x)-f(x) \right)\,dx$ where $a$ and $b$ are the $x$ -coordinates of the points of intersection. Finding the $x$ -coordinates of the intersection points We can find the $x$ -coordinate of each point of intersection by setting the functions equal to each other and solving the resulting equation. $\begin{aligned} f(x)&=g(x) \\\\ x^2+2x &=2x+1\\\\ x^2 &= 1 \\\\ x &= \pm 1 \end{aligned}$ The graphs intersect where $x=-1$ and $x=1$. Setting up the definite integral Thus, the area of the shaded region pictured above is given by: $\begin{aligned} &\phantom{=} \int_{-1}^{1}\left(2x+1-(x^2+2x)\right)\,dx \\\\ &= \int_{-1}^{1}\left(1-x^2\right)\,dx \end{aligned}$ Evaluating the definite integral $\begin{aligned} &\phantom{=} \int_{-1}^{1} \left( 1- x^2 \right) \,dx \\\\ &= x - \dfrac{x^3}{3}~\Bigg|_{-1}^{1} \\\\ &= \left( 1-\dfrac{1}{3} \right) - \left( -1+\dfrac{1}{3} \right) \\\\ &= \dfrac{4}{3} \end{aligned}$ Answer The area is $ \dfrac{4}{3}$ square units.